Posted by **pavatharani** on Wednesday, July 13, 2011 at 11:07am.

a car accelerates from rest at a constant rate 'a' for some time. after which it decelerates at a constant rate'b' and comes to rest. if the total time elapsed is t second, evaluate

1) the maximum velocity reached and

2) the total distance travelled.

the answer for this problem is

1) abt/a+b

2) abt^2/2(a+b)

pls tell me how this answer is arrived at

- physics -
**CANTIUS**, Sunday, July 31, 2011 at 6:58am
Your given answer for (2) can't be correct, it doesn't have units of distance (it has units of acceleration).

- physics -
**CANTIUS**, Sunday, July 31, 2011 at 7:06am
In general, v = v_0 + a t, where a is any acceleration.

Let t1 be the time during the acceleration and t2 be the time during the deceleration and v = the speed at the end of the acceleration.

v - 0 = a t1

0 - v = -b t2

v = a t1 = b t2

This yields the equation

a t1 - b t2 = 0

We also know that

t1 + t2 = t

These are two linear equations in two unknowns. There are several ways of solving these. One way is to solve the first equation for t2 in terms of t1, then plug this into the second equation:

t2 = (a/b)t1

t1 + (a/b)t1 = t

(b + a)/b t1 = t

t1 = b/(a + b) t

Similarly, t2 = a/(a + b) t

From one of the first equations

v = a t1 = a[b/(a + b) t] = abt/(a+b)

As I mentioned in an earlier comment, your answer for (2) cannot be correct.

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